§10.prop2.58ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη.
If an area be contained by a rational straight line and a fifth binomial, the side of the square equal to the area is the irrational straight line called that which produces a rational and a medial area.
χωρίον γὰρ τὸ ΑΓ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων πέμπτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ· λέγω, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη ῥητὸν καὶ μέσον δυναμένη.
For let the area AC be contained by the rational straight line AB and the fifth binomial AD divided into its terms at E, so that the greater term is AE; I say that the side of the square equal to the area AC is the irrational straight line called that which produces a rational and a medial area.
Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρότερον δεδειγμένοις·
For let the same construction be made as before.
φανερὸν δή, ὅτι ἡ τὸ ΑΓ χωρίον δυναμένη ἐστὶν ἡ ΜΞ. δεικτέον δή, ὅτι ἡ ΜΞ ἐστιν ἡ ῥητὸν καὶ μέσον δυναμένη.
It is then manifest that the side of the square equal to the area AC is MX. We must then prove that MX is the straight line which produces a rational and a medial area.
ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΗ τῇ ΗΕ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΑΘ τῷ ΘΕ, τουτέστι τὸ ἀπὸ τῆς ΜΝ τῷ ἀπὸ τῆς ΝΞ· αἱ ΜΝ, ΝΞ ἄρα δυνάμει εἰσὶν ἀσύμμετροι.
For since AH is incommensurable with HE, AT is also incommensurable with TE, that is, the square on MN with the square on NX; therefore MN, NX are incommensurable in square.
καὶ ἐπεὶ ἡ ΑΔ ἐκ δύο ὀνομάτων ἐστὶ πέμπτη, καί ἔλασσον αὐτῆς τμῆμα τὸ ΕΔ, σύμμετρος ἄρα ἡ ΕΔ τῇ ΑΒ μήκει.
And since AD is a fifth binomial, and ED is its lesser segment, therefore ED is commensurable in length with AB.
ἀλλὰ ἡ ΑΕ τῇ ΕΔ ἐστιν ἀσύμμετρος· καὶ ἡ ΑΒ ἄρα τῇ ΑΕ ἐστιν ἀσύμμετρος μήκει.
But AE is incommensurable with ED; therefore AB is also incommensurable in length with AE.
μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, τουτέστι τῇ ΕΚ, ἀλλὰ ἡ ΔΕ τῇ ΕΖ σύμμετρός ἐστιν, καὶ ἡ ΕΖ ἄρα τῇ ΕΚ σύμμετρός ἐστιν.
Therefore AK is medial, that is, the sum of the squares on MN, NX is medial. And since DE is commensurable in length with AB, that is, with EK, while DE is commensurable with EZ, therefore EZ is also commensurable with EK.
καὶ ῥητὴ ἡ ΕΚ· ῥητὸν ἄρα καὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ ΜΝΞ·
And EK is rational; therefore EL is also rational, that is, MP is, that is, the rectangle contained by MN, NX is.
αἱ ΜΝ, ΝΞ ἄρα δυνάμει ἀσύμμετροί εἰσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν.
Therefore MN, NX are incommensurable in square, making the sum of the squares on them medial, but the rectangle contained by them rational.
ἡ ΜΞ ἄρα ῥητὸν καὶ μέσον δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ χωρίον· ὅπερ ἔδει δεῖξαι.
Therefore MX is the straight line which produces a rational and a medial area, and is the side of the square equal to the area AC; which was to be proved.
§10.prop2.59ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν ἡ καλουμένη δύο μέσα δυναμένη.
If an area be contained by a rational straight line and a sixth binomial, the side of the square equal to the area is the irrational straight line called that which produces two medial areas.
χωρίον γὰρ τὸ ΑΒΓΔ περιεχέσθω ὑπὸ ῥητῆς τῆς ΑΒ καὶ τῆς ἐκ δύο ὀνομάτων ἕκτης τῆς ΑΔ διῃρημένης εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον ὄνομα εἶναι τὸ ΑΕ· λέγω, ὅτι ἡ τὸ ΑΓ δυναμένη ἡ δύο μέσα δυναμένη ἐστίν.
For let the area ABCD be contained by the rational straight line AB and the sixth binomial AD divided into its terms at E, so that the greater term is AE; I say that the side of the square equal to the area AC is the straight line which produces two medial areas.
Κατεσκευάσθω τὰ αὐτὰ τοῖς προδεδειγμένοις.
Let the same construction be made as before.
φανερὸν δή, ὅτι τὸ ΑΓ δυναμένη ἐστὶν ἡ ΜΞ, καὶ ὅτι ἀσύμμετρός ἐστι ἡ ΜΝ τῇ ΝΞ δυνάμει.
It is then manifest that the side of the square equal to the area AC is MX, and that MN is incommensurable in square with NX.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΑ τῇ ΑΒ μήκει, αἱ ΕΑ, ΑΒ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
And since EA is incommensurable in length with AB, therefore EA, AB are rational straight lines commensurable in square only.
μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ. πάλιν, ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΕΔ τῇ ΑΒ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΖΕ τῇ ΕΚ·
Therefore AK is medial, that is, the sum of the squares on MN, NX is medial. Again, since ED is incommensurable in length with AB, therefore ZE is also incommensurable with EK.
αἱ ΖΕ, ΕΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
Therefore ZE, EK are rational straight lines commensurable in square only.
μέσον ἄρα ἐστὶ τὸ ΕΛ, τουτέστι τὸ ΜΡ, τουτέστι τὸ ὑπὸ τῶν ΜΝΞ. καὶ ἐπεὶ ἀσύμμετρος ἡ ΑΕ τῇ ΕΖ, καὶ τὸ ΑΚ τῷ ΕΛ ἀσύμμετρόν ἐστιν.
Therefore EL, that is, MP, is medial, that is, the rectangle contained by MN, NX is. And since AE is incommensurable with EZ, AK is also incommensurable with EL.
ἀλλὰ τὸ μὲν ΑΚ ἐστι τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝ, ΝΞ, τὸ δὲ ΕΛ ἐστι τὸ ὑπὸ τῶν ΜΝΞ·
But AK is the sum of the squares on MN, NX, and EL is the rectangle contained by MN, NX; therefore the sum of the squares on MN, NX is incommensurable with the rectangle contained by MN, NX.
ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΜΝΞ τῷ ὑπὸ τῶν ΜΝΞ. καί ἐστι μέσον ἑκάτερον αὐτῶν, καὶ αἱ ΜΝ, ΝΞ δυνάμει εἰσὶν ἀσύμμετροι.
And each of them is medial, and MN, NX are incommensurable in square.
ἡ ΜΞ ἄρα δύο μέσα δυναμένη ἐστὶ καὶ δύναται τὸ ΑΓ· ὅπερ ἔδει δεῖξαι.
Therefore MX is the straight line which produces two medial areas, and is the side of the square equal to the area AC; which was to be proved.